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Conical Pendulum--- did I do this right?
A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string.
1)Determine the Mass of the Ball.
2)Determine the Speed of the Ball.
3)Determine the Frequency of revolutions of the Ball.
4)Suppose that the string breaks as the ball swings in its circular path. Qualitatively describe the trajectory of the ball after the string breaks but before it hits the ground.
1)
F=Tsin(theta)=mg
m= (Tcos theta)/g
2)
Tcos(theta) = mg
T= mg/[cos(theta)]
F=Tsin(theta)=MAc
T=sin(theta)=(mv^2)/r
F= mg/[cos(theta)] * sin(theta) subtitute for T
F=mgtan(theta)
r=Lsin(theta)
F=(mv^2) / Lsin(theta)
mgtan(theta)=(mv^2)/ Lsin(theta)
v= square root of gLtan(theta)sin(theta)
3)
f= 1/t
t= x/v = 2(pi)r/v = 2(pi)Lsin(theta) / square root of gLtan(theta)sin(theta)
f= square root of gLtan(theta)sin(theta) / 2(pi)Lsin(theta)
4) What else can I say besides that it goes out in a straight line tangent to the "circle"?
Homework Statement
A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string.
1)Determine the Mass of the Ball.
2)Determine the Speed of the Ball.
3)Determine the Frequency of revolutions of the Ball.
4)Suppose that the string breaks as the ball swings in its circular path. Qualitatively describe the trajectory of the ball after the string breaks but before it hits the ground.
The Attempt at a Solution
1)
F=Tsin(theta)=mg
m= (Tcos theta)/g
2)
Tcos(theta) = mg
T= mg/[cos(theta)]
F=Tsin(theta)=MAc
T=sin(theta)=(mv^2)/r
F= mg/[cos(theta)] * sin(theta) subtitute for T
F=mgtan(theta)
r=Lsin(theta)
F=(mv^2) / Lsin(theta)
mgtan(theta)=(mv^2)/ Lsin(theta)
v= square root of gLtan(theta)sin(theta)
3)
f= 1/t
t= x/v = 2(pi)r/v = 2(pi)Lsin(theta) / square root of gLtan(theta)sin(theta)
f= square root of gLtan(theta)sin(theta) / 2(pi)Lsin(theta)
4) What else can I say besides that it goes out in a straight line tangent to the "circle"?